There are two functions Rolle f,g:[a, b}->R with g(x)<>0, g'(x)<>0, what ever it is x in the compact interval [a, b] and f(a)/g(a)=f(b)/g(b).
You must demonstrate that there is at least point c in the interval (a, b) so as f(c)/g(c)=f'(c)/g'(c).
We are looking for one function h so as h'(c)=0 and h'(c)=0 so as to give us the relation: f(c)/g(c)=f'(c)/g'(c), which is equivalent with f'(c)*g(c)=f(c)*g'(c)<==>f(c)*g'(c)-f'(c)*g(c)=0
==> h'(c)=f(c)*g'(c)-f'(c)*g(c)==>
h'(x)=f(x)*g'(x)-f'(x)*g(x) we consider h(x)=g(x)/f(x) f, g are functions Rolle ==> they are continue on the interval [a, b] and derivate on interval (a, b)
h(a)=g(a)/h(a) and h(b)=g(b)/h(b)==> h(a)=h(b)==>g(a)/f(a)=g(b)/f(b)
==>f(a)/g(a)=f(b)/g(b)==> there is c in the interval (a, b) so as h'(c)=0
h'(x)=[f(x)*g'(x)-f'(x)*g(x)]/[f(x)*f(x)]
h'(c)=0==>[f(c)*g'(c)-f'(c)*g(c)]/[f(c)*f(c)]=0==>
f(x)*g'(x)-f'(x)*g(x)=0

Written by Florina Leach
Teacher: Ligia Garlea
"Alex. Papiu Ilarian" HS Dej, Cluj, Romania